About $6\times {10}^{-3}$mol $K_{2}C{r}_2{O}_7$ reacts completely with  $9\times {10}^{-3}$mol $X^{n+}$ to give $X{O}_3^{-}$ in acidic medium.  The value of ‘n’ is      

$K_{2}C{r}_2{O}_7+X^{+n}\to C{r}^{+3}+X{O}_3^{-}$

$6\times {10}^{-3}$    $9\times {10}^{-3}$ 

  2         :         3

no.of electron involved = no.of electrons

in oxidation $\left(X^{+n}\right)$        involved in

Reduction$\left(K_{2}C{r}_2{O}_7\right)$          

2(6)      =   3(5-n)

5 – n = 4

$\therefore n=1$